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Describe and correct the error in $\square$nding $\mathrm{m}<\mathrm{A}$ in $\Delta \mathrm{ABCwhen} \mathrm{a}=19, \mathrm{b}=21,$ and $\mathrm{c}=11 .$
$\cos A=\frac{19^{2} \square 21^{2}\square11^{2}}{\square2(19)(21)}$
$m \angle A \approx 75.4 \square$

It's on this fun. We're going to describe the errors that we find. So let's start beginning. So we seem parentheses, which means we're gonna be doing distribution. So we're going to distribute the seven to the sea. The tree. See, They did so that's good. But then they didn't distribute the seven to the negative six. So instead of seven C minus six, it should be seven C minus 42. Okay, so that's on the left side. On the right side, we see more distribution. So we're going to distribute this in so they distribute it to the sea. They distributed the three to the negative, see? So that's good. But again, they didn't distribute it to the second one. So we have this four here that should actually be but 12. So this is what we should have at this point. Then when we put this all together, we're going to add the three c over. So we do end up with the 10 C. So that was right. But then we're also going to add 42 to both sides and we'll end up with a 54 so we can see that number differs from here. Then we'll divide by 10 and we'll get a C equals 54 tents. So our number over here was incorrect. So we found our errors. They did not distribute properly on either side. And then that changed the whole problem, making our solution incorrect. So now this is the

All right. So we're going to see what errors we confined on this problem. So starting at the beginning, we would use distribution, which we can see would be two times for beam. So they got eight B and then two times negative three and they got negative six. So we have a B minus six and then eight B minus six over here as well. Since we have eight b on both sides, it looks like they subtracted eight b from both sides. And that would cancel both a fees. Leaving us with negative six equals negative. Six. So far, we are correct. The final step they do is saying that B equals negative six. This is where we see our issue, since negative six equals negative six. There's no variable there. We can't input any variable. Negative. Six equals negative. Six tells us this is an identity and B can be any number. Any number that you plug in for B. We'll get you with negative six equals negative. Six. So the error was saying that B is negative. Six when in fact, it is just negative six equal to negative six. Meaning

So this is what is given. We have minus to be cubed plus 12 b squared minus 14 B. That becomes minus to be. So let's say we factor out negative to be were left with B squared. That is correct. But what's incorrect here that this should be minus and then this should be Plus, so this is incorrect here. So let's move on from there. So minus to be into B squared minus six. B plus seven? No. Are there any two values that would give a product of seven and a sum of negative six? There are none. If we choose positive one and positive seven. That would give us some of eight negative seven and negative one that also give negative eight, So this should be your final answer here.

So in question 79 we're supposed to find what's wrong with how they solved the limit. They took this limit over here and they just took the derivative of the top. The use capitals rule over the derivative of the bottom. And at first it seems there's nothing wrong with that. But you know, the first thing I would do is I would check if there was an indeterminant form. So if it if it doesn't have the right in determining form, you can't use low petals role. So what you need to do now, we're going to do direct substitution. So it'll be I'm gonna plug into into this original limit here. So three times two squared plus four times two a swan, all divided by two squared minus two minus two. So that will make the top That will be that'll be 12 plus eight plus one, which is 21/0. And yes, this is an indeterminate form. But the ones that were looking for when doing low petals rule are 0/0 and plus or minus infinity over infinity. So this one does not qualify no good. And so so you were not allowed to use literal and solving this limit. And that is why this was done incorrectly.